Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Gamma distribution ▷ Kullback-Leibler divergence

Theorem: Let $X$ be a random variable. Assume two gamma distributions $P$ and $Q$ specifying the probability distribution of $X$ as

$\label{eq:gams} \begin{split} P: \; X &\sim \mathrm{Gam}(a_1, b_1) \\ Q: \; X &\sim \mathrm{Gam}(a_2, b_2) \; . \\ \end{split}$

Then, the Kullback-Leibler divergence of $P$ from $Q$ is given by

$\label{eq:gam-KL} \mathrm{KL}[P\,||\,Q] = a_2 \, \ln \frac{b_1}{b_2} - \ln \frac{\Gamma(a_1)}{\Gamma(a_2)} + (a_1 - a_2) \, \psi(a_1) - (b_1 - b_2) \, \frac{a_1}{b_1} \; .$

Proof: The KL divergence for a continuous random variable is given by

$\label{eq:KL-cont} \mathrm{KL}[P\,||\,Q] = \int_{\mathcal{X}} p(x) \, \ln \frac{p(x)}{q(x)} \, \mathrm{d}x$

which, applied to the gamma distributions in \eqref{eq:gams}, yields

$\label{eq:gam-KL-s1} \begin{split} \mathrm{KL}[P\,||\,Q] &= \int_{-\infty}^{+\infty} \mathrm{Gam}(x; a_1, b_1) \, \ln \frac{\mathrm{Gam}(x; a_1, b_1)}{\mathrm{Gam}(x; a_2, b_2)} \, \mathrm{d}x \\ &= \left\langle \ln \frac{\mathrm{Gam}(x; a_1, b_1)}{\mathrm{Gam}(x; a_2, b_2)} \right\rangle_{p(x)} \; . \end{split}$

Using the probability density function of the gamma distribution, this becomes:

$\label{eq:gam-KL-s2} \begin{split} \mathrm{KL}[P\,||\,Q] &= \left\langle \ln \frac{ \frac{ {b_1}^{a_1}}{\Gamma(a_1)} x^{a_1-1} \exp[-b_1 x] }{ \frac{ {b_2}^{a_2}}{\Gamma(a_2)} x^{a_2-1} \exp[-b_2 x] } \right\rangle_{p(x)} \\ &= \left\langle \ln \left( \frac{ {b_1}^{a_1}}{ {b_2}^{a_2}} \cdot \frac{\Gamma(a_2)}{\Gamma(a_1)} \cdot x^{a_1-a_2} \cdot \exp[-(b_1-b_2) x] \right) \right\rangle_{p(x)} \\ &= \left\langle a_1 \cdot \ln b_1 - a_2 \cdot \ln b_2 - \ln \Gamma(a_1) + \ln \Gamma(a_2) + (a_1-a_2) \cdot \ln x - (b_1-b_2) \cdot x \right\rangle_{p(x)} \; . \end{split}$ $\label{eq:gam-means} \begin{split} x \sim \mathrm{Gam}(a,b) \quad \Rightarrow \quad &\left\langle x \right\rangle = \frac{a}{b} \quad \text{and} \\ &\left\langle \ln x \right\rangle = \psi(a) - \ln(b) \; , \end{split}$

the Kullback-Leibler divergence from \eqref{eq:gam-KL-s2} becomes:

$\label{eq:gam-KL-s3} \begin{split} \mathrm{KL}[P\,||\,Q] &= a_1 \cdot \ln b_1 - a_2 \cdot \ln b_2 - \ln \Gamma(a_1) + \ln \Gamma(a_2) + (a_1-a_2) \cdot \left( \psi(a_1) - \ln(b_1) \right) - (b_1-b_2) \cdot \frac{a_1}{b_1} \\ &= a_2 \cdot \ln b_1 - a_2 \cdot \ln b_2 - \ln \Gamma(a_1) + \ln \Gamma(a_2) + (a_1-a_2) \cdot \psi(a_1) - (b_1-b_2) \cdot \frac{a_1}{b_1} \; . \end{split}$

Finally, combining the logarithms, we get:

$\label{eq:gam-KL-qed} \mathrm{KL}[P\,||\,Q] = a_2 \, \ln \frac{b_1}{b_2} - \ln \frac{\Gamma(a_1)}{\Gamma(a_2)} + (a_1 - a_2) \, \psi(a_1) - (b_1 - b_2) \, \frac{a_1}{b_1} \; .$
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Metadata: ID: P93 | shortcut: gam-kl | author: JoramSoch | date: 2020-05-05, 08:41.